In Rotman he defined the Augmented singular complex by extending the singular chain complex of a space $$...\rightarrow S_2(X) \rightarrow S_1(X) \rightarrow S_0(X) \rightarrow 0$$ By defining $$\epsilon (\sum m_x x)= \sum m_x[\emptyset]$$.

In spanner he says this map $$\epsilon$$ must be subjective hence, possibility of considering $$X=\emptyset$$ goes away.

Hatcher says, we should choose $$X$$ to be nonempty to avoid getting nonzero homology groups of negative degree.

But Rotman never mentions anything about emptiness of $$X$$.

Later he gives a problem which says,

If $$A\subset X$$ , then there is an exact sequence $$…\rightarrow \tilde H_n(A)\rightarrow \tilde H_n(X) \rightarrow H_n(X,A)\rightarrow …$$ , which ends at

$$…\rightarrow \tilde H_0(A) \rightarrow \tilde H_0(X) \rightarrow H_0(X,A) \rightarrow 0$$

For, $$A\neq \emptyset$$ the problem is plain coming from the equality of chain complexes , $$\tilde S_*(X)/\tilde S_*(A) = S_*(X)/S_*(A)$$

If I put $$A=\emptyset$$ then depending upon $$X$$ to be non empty or empty various cases are coming and some of them are contradictory.

For example, if $$A=\emptyset$$ and $$X\neq \emptyset$$ then from the exact sequence $$\tilde H_0(A)=0, H_0(X,A)= H_0(X)$$ so, $$\tilde H_0(X)\cong H_0(X)$$, but this contradicts the relation of 0th reduced homology and 0th homology group in terms of rank, when X has finitely many path components.

Is there any way to deal with this?

Does the usual practice of reduced homology groups deal with nonempty spaces only?

• That's a good question, this convention can be tricky. You should try to see what happens if you always put a $\mathbb Z$ as $S_{-1}(X)$, even if $X$ is empty (thus allowing one negative homology group for emptyspaces). May 9 '20 at 8:05

The augmented chain complex of a space $$X$$ is $$...\rightarrow S_2(X)\stackrel{\partial}{\rightarrow} S_1(X) \stackrel{\partial}{\rightarrow} S_0(X) \stackrel{\epsilon}{\rightarrow} \mathbb Z \to 0 .$$ This is defined also for $$X = \emptyset$$, but in fact is usually only considered for non-empty $$X$$. The reduced homology groups of $$X$$ are the homology groups of the augmented chain complex, therefore we have $$\tilde H_n(X) = H_n(X)$$ for $$n > 0$$. For $$n = 0$$ we get $$\tilde H_0(X) = \ker(\epsilon)/\text{im}(\partial)$$ which can be identified with a subgroup of $$H_0(X) = S_0(X)/\text{im}(\partial)$$. Moreover, one can easily show that $$\tilde H_0(X) \approx \ker(p_* : H_0(X) \to H_0(*))$$, where $$p : X \to *$$ is the unique map to a one-point space $$*$$. Note that $$\tilde H_0(X) = 0$$ for $$X = \emptyset$$.
What happens for $$n = -1$$? If $$X \ne \emptyset$$, then $$\epsilon$$ is surjective and $$\tilde H_{-1}(X) = 0$$, but if $$X = \emptyset$$, then we get $$S_0(X) = 0$$ and $$\tilde H_{-1}(X) = \mathbb Z$$. This is why Hatcher says that we should choose $$X$$ to be non-empty to avoid getting nonzero reduced homology groups of negative degree. However, it is no real problem to allow also $$X = \emptyset$$.
Without assuming $$A \ne \emptyset$$, the exact sequence of $$(X,A)$$ $$…\rightarrow \tilde H_n(A)\rightarrow \tilde H_n(X) \rightarrow H_n(X,A)\rightarrow …$$ ends with $$…\rightarrow \tilde H_0(A) \rightarrow \tilde H_0(X) \rightarrow H_0(X,A) \rightarrow \tilde H_{-1}(A) \rightarrow \tilde H_{-1}(X) \to 0$$ You see that for $$A \ne \emptyset$$ we get $$\tilde H_{-1}(A) = \tilde H_{-1}(X) = 0$$ which yields Rotman's and Hatcher's sequence. For $$A = \emptyset$$ we get $$…\rightarrow \tilde H_0(A) \rightarrow \tilde H_0(X) \rightarrow H_0(X,A) \rightarrow \mathbb Z \rightarrow \tilde H_{-1}(X) \to 0$$ where $$\tilde H_{-1}(X) = 0$$ if $$X \ne \emptyset$$ and $$\tilde H_{-1}(X) = \mathbb Z$$ if $$X = \emptyset$$.